Chapter 5: Problem 168
On mixing, heptane and octane form an ideal solution. At \(373 \mathrm{~K}\), the vapour pressures of the two liquid components (heptane and octane) are \(105 \mathrm{kPa}\) and \(45 \mathrm{kPa}\) respectively. Vapour pressure of the solution obtained by mixing \(25.0 \mathrm{~g}\) of heptane and \(35 \mathrm{~g}\) of octane will be (molar mass of heptane \(=100 \mathrm{~g} \mathrm{~mol}^{-1}\) and of octane \(=\) \(114 \mathrm{~g} \mathrm{~mol}^{-1}\) ) (a) \(72.0 \mathrm{kPa}\) (b) \(36.1 \mathrm{kPa}\) (c) \(96.2 \mathrm{kPa}\) (d) \(144.5 \mathrm{kPa}\)
Short Answer
Step by step solution
Key Concepts
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